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Moment of inertia of a hollow sphere proof

  • #1
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I'm wondering how you would prove the (2/3)MR^2 moment of inertia of a hollow sphere.
My idea was to brake it into like.......sort of cylindrical shells. ones with that are very very short, like little wires making a hollow sphere.
I=int[(r^2)dm]
I=[M/(4pi(R^2))]int[(R^2-y^2)(2pi(R^2-y^2)^.5)dy]
I=[M/(2R^2)]int[(R^2-y^2)^(3/2)dy]
I am just in high school so I wont follow if you guys do anything like a double integral, so if there is a way around that, tell me that way. Thanks!
 

Answers and Replies

  • #2
OlderDan
Science Advisor
Homework Helper
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I'm wondering how you would prove the (2/3)MR^2 moment of inertia of a hollow sphere.
My idea was to brake it into like.......sort of cylindrical shells. ones with that are very very short, like little wires making a hollow sphere.
I=int[(r^2)dm]
I=[M/(4pi(R^2))]int[(R^2-y^2)(2pi(R^2-y^2)^.5)dy]
I=[M/(2R^2)]int[(R^2-y^2)^(3/2)dy]
I am just in high school so I wont follow if you guys do anything like a double integral, so if there is a way around that, tell me that way. Thanks!
You can break it into a stack of rings. I assume that is what you are calling very short cylinders. The moment of inertia of each ring is (dm)r². The mass (dm) of each ring depends on the radius of the ring in a way that is a bit complicated if you integrate along an axis through the center of the sphere. It is less complicated if you integrate over an angle from the center of the sphere, but you then have to express r in terms of the angle. Either way will get you the answer.
 
  • #3
225
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somehow I am a little bit scared to try the angle idea. Do you mean that considering @ to be the angle with the negative verticle, I add up all of the (dm)(Rcos@)^2 from 0 to 180?
If you look at my work, I do know how to set r as a function of y, but something is clearly wrong with my work...
(oh by the way, thanks because you helped me a little with the solid sphere earlier)
 
  • #4
225
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I am trying the angle method, but looking ahead, I'm not really sure how I would get the intergral to have a d@ in it. I'll work on it.
it there such a thing as angular density???!
 
  • #5
225
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Well, I tried but it is wrong:
I=int[(dm)Rcos@]
I=(sigma)integral from 0 to pi[(ds)2pi(Rcos@)^2]
ds=R(d@)
I=[MR/2][.5pi+.25sin(2pi)]
I=(MRpi)/4

***sigma is surface density
 
  • #6
OlderDan
Science Advisor
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Well, I tried but it is wrong:
I=int[(dm)Rcos@]
I=(sigma)integral from 0 to pi[(ds)2pi(Rcos@)^2]
ds=R(d@)
I=[MR/2][.5pi+.25sin(2pi)]
I=(MRpi)/4

***sigma is surface density
I think you may be mixing up sines and cosines, and maybe lost a square. How are you defining your angle? It looks like you want to intergrate from zero to pi. In that case your first equation should be

I=int[(dm)R²sin²@]

where Rsin@ is the radius of the ring. If the radius of the ring is Rcos@, then the limits of integration are from -pi/2 to pi/2.

The surface area associated with a ring is Rd@*(2pi)*(radius of ring) using the expression for the radius that fits your definition of the angle. I'm seeing a factor of (2pi)R^4 and the integral of the cube of either sin@ or cos@ depending on your angle definition.
 
  • #7
PSOA
I am in high school too. How do you get rid of dm?

Thanks in advance.
 
  • #8
OlderDan
Science Advisor
Homework Helper
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I am in high school too. How do you get rid of dm?

Thanks in advance.
You integrate over dm by first expressing it in terms of the area of the sphere. There is a constant mass per unit area defined by

σ = M/A = M/(4piR²)

A small area of the sphere then has mass given by

dm = σdA

dA depends on how you define your little pieces of area. In this problem the apoproach is to define dA to be the area of a thin slide of the sphere perpendicular to a diameter.

I must tell you that your posting of the question out of context of the problem in a new thread is complete nonsense. There is no answer to your question. You don't "get rid of dm". You add together all the products of the form

dI = (dm)r²

to find the total moment of inertia of the spherical shell.
 
  • #9
225
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wow I did mix up sine and cosine--that's embarrassing :O
thanks a lot!
 
  • #10
225
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got it! thanks for the help again
 
  • #11
I proved the Moment of Inertia of a Hollow Sphere = (2/3)*M*r^2 by using double intigration (Area Intigral) method.

Please see the attachement for, how to prove it.
 

Attachments

  • #12
1
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can u write the provement of the formula clearly?
 
  • #13
2
0
hey pal_physics, thanx 4 solution, i quite knew the subject but i was in a hurry doing one assignment- so very helpful. I'm wondering what keeps people like you motivated to this voluntary work
 
  • #14
1
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hey. there is a "trick" here as usual with solutions.
moment of each hoop I = r^2*dm
area of sphere A=4piR^2

heres the trick. you want dA at r = R, so
dA = 8pi*rdr --> dA = 8pi*Rdr

dM = MdA/A = MdA/4piR^2

dI=r^2*M*dA/(4piR^2)
dI=(M/4piR^2)*(r^2)8pi*Rdr
dI=2M/R * r^2dr
I = 2M/R * R^3/3
I = 2MR^2/3
 
  • #15
25
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Is there anyway to prove the moment of inertia of a hollow sphere by using the y^2 + x^2 = r^2 method instead of the angle method?
 

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